About Question enthuware.ocajp.i.v7.2.1189 :

[stuie382]

Ok so I'm pretty sure the question asked me to select 3 answers, however during the review it shows that I needed 4 answers.

[Shortrope]

I got this question correct but don't understand why opt 4 the description says

Since Runnable is an interface, it cannot be instantiated like this. But you can do :
Runnable r = new Runnable(){
public void run(){
}                     
};

So you can instantiate an interface? But an interface is abstract.

[admin]

You are not really instantiating an interface here. You are instantiating an anonymous class that implements the interface.
Not required for the exam, but you may go through this if you would like to know more: http://docs.oracle.com/javase/tutorial/ ... asses.html

HTH,
Paul.

[crazymind]

Hi, why does char ch = 10; works without any casting? I understand it can fit in a char but you always need a cast when you assign a int to a char, right ?

[crazymind]

Ok, I see, it is narrow primitive conversion.

[admin]

More precisely "implicit narrowing" i.e. narrowing without any explicit cast. Explained on pg 69 section 3.3.3 of OCAJP 8 Fundamentals by Hanumant Deshmukh book.

[enthunoob]

char ch =10; works because even though it is an int, it is a compile time constant, thus the compiler can figure out it is safe.

Examples from the above references book (page 73):
char c3 = 1; //ok, even though 1 is an int but it is a compile time constant whose value can fit into a char.
char c4 = -1; //will not compile because -1 cannot fit into a char